Solve the logarithmic equation. Be sure to reject any value of \( x \) that is not in the domain of the original logarithmic expressio \[ \log _{3}(x+7)+\log _{3}(x+5)=1 \] Solve the equation. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The solution set is \( \square \) 3. (Simplify your answer. Use a comma to separate answers as needed.) B. There are infinitely many folutions. C. There is no solution.

To solve the equation \(\log_{3}(x+7) + \log_{3}(x+5) = 1\), we can use the property of logarithms that states \(\log_{a}(b) + \log_{a}(c) = \log_{a}(bc)\). Therefore, we can rewrite the equation as: \[ \log_{3}((x+7)(x+5)) = 1 \] Next, we can rewrite the logarithmic equation in exponential form: \[ (x+7)(x+5) = 3^1 \] This simplifies to: \[ (x+7)(x+5) = 3 \] Expanding the left-hand side, we get: \[ x^2 + 12x + 35 = 3 \] Subtracting 3 from both sides yields: \[ x^2 + 12x + 32 = 0 \] Next, we can factor the quadratic equation: \[ (x + 4)(x + 8) = 0 \] Setting each factor equal to zero gives us the solutions: \[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \] \[ x + 8 = 0 \quad \Rightarrow \quad x = -8 \] We need to check whether these solutions fall within the domain of the original logarithmic expressions. For \(\log_{3}(x+7)\) and \(\log_{3}(x+5)\) to be defined, both \(x + 7 > 0\) and \(x + 5 > 0\) must hold. 1. For \( x = -4 \): - \( x + 7 = -4 + 7 = 3 \) (valid) - \( x + 5 = -4 + 5 = 1 \) (valid) 2. For \( x = -8 \): - \( x + 7 = -8 + 7 = -1 \) (not valid) - \( x + 5 = -8 + 5 = -3 \) (not valid) Since only \(x = -4\) is valid, we conclude that the solution set is \(-4\). Thus, the correct choice is: A. The solution set is \(-4\).